A laboratory with a single instrument: a movable triangle. We drag the vertices, notice what doesn’t change, and prove why, from the congruence tests to the Euler line.
which three pieces of data determine a triangle uniquely and which don’t, and how that’s proved
find the angles, sides and area of a triangle: the angle sum, the triangle inequality, similarity, Heron’s formula
where the medians, bisectors, perpendicular bisectors and altitudes meet, and why
The last chapter ended with the fact that any polygon can be cut into triangles by diagonals. There’s a second reason geometry starts with the triangle. Nail four slats into a frame and it skews at the slightest push. A frame of three slats stands rock solid. That’s why bridge trusses, crane jibs and power-line pylons are made of triangles: the triangle is the only polygon of hinged rods that can’t change shape.
Rigidity means that three lengths already determine the whole triangle. What other sets of data determine it? A triangle has six numbers, three sides and three angles, and the question is how many of them, and which ones, you need to know. This chapter is set up as a laboratory with one instrument: a movable triangle. In each experiment we first drag the vertices and notice what stays the same, and then prove why it will be so for every triangle, not just the ones that happened to turn up. The proofs move too: the drawing is built step by step along with the text, and you can drag the vertices at any step.
Six numbers, and the three you need
Experiment 1. Choose what’s known about the triangle and move the sliders. The instrument draws every triangle that fits the description.
Six sets of data. For each, check: can there be two triangles? None? Infinitely many?
Here’s the picture. Three sides, two sides and the angle between them, a side and two angles: in all these cases you get one triangle or none at all. Three angles don’t fix the size: a whole family of triangles of the same shape fits. And two sides with an angle that isn’t between them sometimes describe two different triangles at once.
Congruent triangles are triangles that can be made to coincide by laying one on the other. All three sides and all three angles of one equal the corresponding sides and angles of the other.
The sets of data that give exactly one triangle are called congruence tests. There are three of them, and they have to be proved in order: each proof relies on the ones before.
If two sides and the angle between them in one triangle are equal to two sides and the angle between them in another, the triangles are congruent.
The trick is to lay one triangle on the other and follow where the vertices land.
Let $AB = A'B'$, $AC = A'C'$ and $\angle A = \angle A'$. On the drawing equal sides carry the same tick marks, and equal angles the same colour.Move triangle $ABC$ so that vertex $A$ lands on $A'$ and the ray $AB$ runs along the ray $A'B'$. The segments $AB$ and $A'B'$ are equal and laid off from the same point along the same ray, so $B$ lands exactly on $B'$.Angle $A$ equals angle $A'$, so the ray $AC$ runs along the ray $A'C'$ (if vertex $C$ ended up on the other side of $A'B'$, flip the triangle first). On this ray $AC = A'C'$, so $C$ lands on $C'$.All three vertices coincide, and only one line passes through two points, so the sides $BC$ and $B'C'$ coincide too. The triangles have been made to coincide: they’re congruent.
This is Euclid’s proof (proposition I.4), and in the last chapter we saw its weakness: the postulates don’t allow figures to be “moved”. Hilbert did the honest thing and took as an axiom of congruence a statement from which the first test follows at once. So the first test expresses a property of the plane itself: a triangle can be moved anywhere and it won’t change.
If a side and the two angles on it in one triangle are equal to a side and the two angles on it in another, the triangles are congruent.
The third vertex is the point where two rays meet, and two rays on different lines meet in at most one point.
Let $AB = A'B'$, $\angle A = \angle A'$ and $\angle B = \angle B'$.The vertex $C'$ lies on the ray from $A'$ at the angle $\angle A$ to $A'B'$, and on the ray from $B'$ at the angle $\angle B$. These rays have no other common point: two different lines can’t meet twice, or there would be two lines through two points.Lay off on the first ray the segment $A'C'' = AC$.Triangles $ABC$ and $A'B'C''$ are congruent by the first test: $AB = A'B'$, $AC = A'C''$, $\angle A = \angle A'$. So $\angle A'B'C'' = \angle ABC = \angle A'B'C'$, and the ray $B'C''$ coincides with the second ray.The point $C''$ lies on both rays, so $C'' = C'$. Hence $A'C' = AC$, and by the first test triangles $ABC$ and $A'B'C'$ are congruent.
For the third test we’ll need a property of the isosceles triangle, the pons asinorum or “bridge of asses”, as schoolboys called it, where weak pupils got stuck.
In an isosceles triangle the base angles are equal. The bisector of the apex angle bisects the base and is perpendicular to it.
Cut the triangle in two with the bisector and prove the halves are congruent.
Let the lateral sides of triangle $ABC$ be equal: $CA = CB$.Draw the bisector $CD$ of angle $C$ (it can be constructed with compass and straightedge: chapter 16). It divides angle $C$ into two equal angles.Triangles $ACD$ and $BCD$ are congruent by the first test: $CA = CB$, the side $CD$ is shared, and the angles at vertex $C$ are equal.Congruent triangles have equal corresponding parts: $\angle A = \angle B$, $AD = DB$ and $\angle ADC = \angle BDC$. The last two angles are adjacent and together make $180^\circ$, so each is a right angle: $CD \perp AB$.
If the three sides of one triangle are equal to the three sides of another, the triangles are congruent.
Fit the second triangle against the first and find two isosceles triangles in the resulting figure: they’ll show that the angles are equal too.
Let $AB = A'B'$, $AC = A'C'$, $BC = B'C'$. On the other side of the ray $AB$ lay off an angle equal to $\angle A'$, and on its arm the segment $AC'' = A'C'$. By the first test triangle $ABC''$ is congruent to $A'B'C'$, so $BC'' = B'C' = BC$. So the second triangle has been “fitted” to the side $AB$ from below, with $AC'' = AC$ and $BC'' = BC$.Join $C$ and $C''$. Suppose the segment $CC''$ crosses $AB$ at an interior point, which happens when angles $A$ and $B$ are acute; we’ll deal with the other cases after the proof.Triangle $ACC''$ is isosceles ($AC = AC''$), so its base angles are equal: $\angle ACC'' = \angle AC''C$.Triangle $BCC''$ is isosceles too ($BC = BC''$), and $\angle BCC'' = \angle BC''C$.Add: $\angle ACB = \angle ACC'' + \angle C''CB = \angle AC''C + \angle CC''B = \angle AC''B$. Triangles $ABC$ and $ABC''$ are congruent by the first test: $AC = AC''$, $BC = BC''$, and the angles between them are equal. And $ABC''$ is congruent to $A'B'C'$, so $ABC$ is congruent to $A'B'C'$.
If the segment $CC''$ passes through vertex $A$ (angle $A$ is right), the points $C$, $A$, $C''$ lie on one line, triangle $CBC''$ is isosceles, and the equality of the angles at $C$ and $C''$ is immediate. If the segment $CC''$ misses $AB$ (angle $A$ is obtuse), the angles aren’t added but subtracted: $\angle ACB = \angle BCC'' - \angle ACC''$. The rest of the argument is the same. The third test also explains the rigidity of the frame: the lengths of three slats already determine all the angles.
But “two sides and an angle not between them” isn’t a test. The instrument shows why: the side $a$ laid off from the point $B$ can reach the ray from $A$ in two places. When this data does determine the triangle after all, we’ll find out in the section on the triangle inequality: to answer, we need to be able to compare sides and angles.
Triangles $ABC$ and $A'B'C'$ have equal sides $AB = A'B' = 6$, $BC = B'C' = 4$ and equal angles $A = A' = 35^\circ$. Are the triangles congruent?
Angle $A$ isn’t between the sides $AB$ and $BC$. The circle of radius $4$ centred at $B$ crosses the ray from $A$ at two points: we get two different triangles with the same data. This is exactly the case the instrument shows in the “two sides and an angle not between them” mode. The triangles may coincide, or they may not.
One hundred and eighty degrees
Experiment 2. Drag the vertices and watch the sum of the angles. Then press the button: the angles at $A$ and $B$ “tear off” and join up at vertex $C$.
Put together, the three angles make a straight angle. The dashed line is the line through $C$ parallel to $AB$.
You can do this experiment with a paper triangle too: tear off the three corners and put them side by side. They always make a straight angle. But paper, like a protractor, only confirms what we see. The proof works just like the animation.
The angles of any triangle add up to $180^\circ$.
The angle at vertex $A$; in the proof’s drawing, $\alpha$.The angle at vertex $B$; in the drawing, $\beta$.The angle at vertex $C$; in the drawing, $\gamma$.Example: if $\angle A = 50^\circ$ and $\angle B = 70^\circ$, then $\angle C = 180^\circ - 50^\circ - 70^\circ = 60^\circ$.
The trick is to gather all three angles at one vertex.
Draw through vertex $C$ the line parallel to side $AB$. Such a line exists, and there’s only one: that’s the fifth postulate in Playfair’s form (chapter 16).The angle between this line and side $CA$ equals angle $A$: they’re alternate angles for the parallel lines and the transversal $AC$. The equality of alternate angles for parallel lines is proposition I.29, the first place Euclid uses the fifth postulate.For the same reason the angle between the line and side $CB$ equals angle $B$: alternate angles for the transversal $BC$.Between them is angle $C$. The three angles at vertex $C$ together fill a straight angle, and that’s $180^\circ$. So $\angle A + \angle B + \angle C = 180^\circ$.
Since the fifth postulate is at work in the proof, the angle sum of a triangle isn’t a property of triangles in general but of Euclid’s plane. On a sphere, where great circles play the role of lines, the angles of a triangle always add up to more than $180^\circ$: a triangle with one vertex at the pole and two on the equator a quarter of the way round from each other has three right angles. That’s chapter 58.
An exterior angle of a triangle equals the sum of the two interior angles not adjacent to it.
The same parallel line, only drawn inside the exterior angle. This is how Euclid argues in proposition I.32.
Extend side $BC$ beyond vertex $C$ to a point $D$. The angle $ACD$ is the exterior angle of the triangle at vertex $C$.Draw from $C$ the ray $CE$ parallel to $BA$ and pointing the same way.$\angle ACE = \angle A$: these are alternate angles for the parallels $BA$, $CE$ and the transversal $AC$.$\angle ECD = \angle B$: these are corresponding angles for the same parallels and the transversal $BD$. Corresponding angles are equal because each equals the same alternate angle, via vertical angles.The ray $CE$ divides the exterior angle into these two parts, so $\angle ACD = \angle A + \angle B$.
The angle sum also gives the congruence test we were missing: a side and two angles, not necessarily on that side.
If a side and two angles of one triangle are equal to a side and two angles of another, and the angles are placed the same way relative to the side, the triangles are congruent. In particular, right triangles are congruent if they have equal hypotenuses and an equal acute angle.
The third angles of the triangles are equal too: each is $180^\circ$ minus the sum of the two given angles. So both angles on the given side are known and equal, and the triangles are congruent by the second test. For right triangles with equal hypotenuses and acute angles it’s exactly the same case: the hypotenuse is the given side, and the two angles are the right one and the acute one.
The angles of a triangle are in the ratio $2 : 3 : 4$. Find the largest angle in degrees.
Let the angles be $2x$, $3x$ and $4x$. Then $2x + 3x + 4x = 9x = 180^\circ$, $x = 20^\circ$, and the largest angle is $4 \cdot 20^\circ = 80^\circ$.
Sticks and the triangle inequality
Experiment 3. Three sticks: the base is fixed, the other two turn on hinges. When do their ends meet?
Change the lengths of the sticks and watch the three inequalities under the drawing.
The triangle inequality is the statement that in any triangle each side is less than the sum of the other two.
To prove it we’ll need a lemma that Euclid proves just before it: it links the lengths of the sides to the size of the opposite angles.
In a triangle the larger side is opposite the larger angle, and conversely, the larger angle is opposite the larger side.
Lay off on the larger side a segment equal to the smaller one, and compare the angles through the resulting isosceles triangle.
Let side $AC$ of triangle $ABC$ be longer than $AB$. Lay off on $AC$ the segment $AD = AB$; since $AB < AC$, the point $D$ lies between $A$ and $C$.Triangle $ABD$ is isosceles, so its base angles at $BD$ are equal: $\angle ABD = \angle ADB$.The angle $ADB$ is the exterior angle of triangle $BDC$ at vertex $D$. By the exterior angle theorem it equals $\angle DBC + \angle C$ and so is greater than angle $C$.The ray $BD$ runs inside angle $ABC$, so $\angle ABC > \angle ABD$. We get the chain $\angle B > \angle ABD = \angle ADB > \angle C$: the larger angle $B$ is opposite the larger side $AC$.The converse we prove by contradiction. Let $\angle B > \angle C$, but side $AC$ not be longer than $AB$. If $AC = AB$, the triangle is isosceles and $\angle B = \angle C$. If $AC < AB$, then by what we’ve proved $\angle B < \angle C$. Both cases contradict the assumption, so $AC > AB$.
In any triangle each side is less than the sum of the other two: $AB < AC + CB$.
Straighten the broken line $ACB$ into a segment and compare it with $AB$ using the lemma.
Extend side $AC$ beyond the point $C$ and lay off $CD = CB$. Then $AD = AC + CB$: the length of the broken line $ACB$ straightened out into a segment.Triangle $BCD$ is isosceles, so $\angle CBD = \angle CDB$.The ray $BC$ runs inside angle $ABD$, so $\angle ABD > \angle CBD = \angle ADB$.In triangle $ABD$ the angle at vertex $B$ is larger than the angle at $D$, and the larger side is opposite the larger angle. So $AD > AB$, that is, $AB < AC + CB$.
You don’t need to check all three inequalities: it’s enough to make sure the longest side is less than the sum of the two shorter ones; the other two checks then hold automatically. If the sum exactly equals the third side, the sticks meet but lie on one line: the triangle flattens into a segment.
This is proposition I.20 of the Elements. Proclus writes that the Epicureans laughed at it: even a donkey knows this, going straight to the hay rather than along two sides. They were wrong to laugh. Euclid wasn’t proving the obvious; he was proving that the obvious follows from the axioms. Besides, the triangle inequality will come in handy again and again: in complex numbers it reads $|z + w| \le |z| + |w|$, and in analysis the very notion of distance rests on it.
The sticks experiment suggests the converse too: if the inequalities hold, the sticks always meet.
If each of three segments $a$, $b$, $c$ is less than the sum of the other two, a triangle can be made from them, and all such triangles are congruent.
The third vertex is a common point of two circles. We need to prove there is one.
Lay off the segment $AB = c$ and draw two circles: centred at $A$ with radius $b$, and centred at $B$ with radius $a$. If they have a common point $C$, then $AC = b$ and $BC = a$: that’s the triangle we want.Take the point $X$ of the second circle that lies on the ray $BA$. Its distance to $A$ is $|c - a|$, and that’s less than $b$: the inequalities $c < a + b$ and $a < b + c$ mean that $c - a < b$ and $a - c < b$. So $X$ lies inside the first circle.The point $Y$ of the second circle on the extension of $AB$ beyond $B$ is at distance $c + a$ from $A$, and that’s more than $b$, since $b < a + c$. The point $Y$ lies outside the first circle.Walk along the second circle from $X$ to $Y$. The distance to $A$ grows continuously from $|c - a|$ to $c + a$ and somewhere equals exactly $b$. At that point $C$ the circles meet. The axiom of continuity is at work here, the very one Euclid was missing (chapter 16).Triangle $ABC$ has sides $a$, $b$ and $c$. Any two triangles with these sides are congruent by the third test.
Two sides of a triangle are $3$ and $8$. How many whole-number values can the length of the third side take?
By the triangle inequality $8 - 3 < c < 8 + 3$, that is, $5 < c < 11$. The whole-number values are $6, 7, 8, 9, 10$: five in all.
Now we can go back to experiment 1 and find out when two sides and an angle not between them do determine a triangle.
Let triangles $ABC$ and $A'B'C'$ have equal sides $AB = A'B' = c$, $BC = B'C' = a$ and equal angles $A = A'$, with $a \ge c$: the angle is opposite the larger of the two sides. Then the triangles are congruent.
If there were two fitting triangles, the side opposite the given angle would turn out shorter than the other given side.
Build both triangles on the same angle $A$ with the same side $AB = c$. Their third vertices $C_1$ and $C_2$ lie on the same ray from $A$ and are at distance $a$ from $B$. Suppose these are different points and $C_1$ is closer to $A$.Triangle $BC_1C_2$ is isosceles, and its base angles at $C_1C_2$ are equal. Both are acute: two angles of a triangle add up to less than $180^\circ$.The angle $AC_1B$ is adjacent to the acute angle $BC_1C_2$, so it’s obtuse.In triangle $ABC_1$ the obtuse angle is the largest, and the largest side is opposite it (the lemma above): $AB > BC_1$, that is, $c > a$.So when $a \ge c$ there can’t be two different points: $C_1 = C_2$, and the triangles coincide.
The most common case of this test is right triangles with equal hypotenuses and an equal leg: the right angle is the largest, so the hypotenuse is longer than a leg. This case has an elegant proof of its own too.
If the hypotenuse and a leg of one right triangle are equal to the hypotenuse and a leg of another, the triangles are congruent.
Put the triangles together along the equal legs: together they make an isosceles triangle.
Let triangles $ABC$ and $A'B'C'$ have right angles at $C$ and $C'$, $AB = A'B'$ and $BC = B'C'$.On the other side of $BC$ lay off the right angle $BCA''$ with $CA'' = C'A'$. By the first test triangle $A''BC$ is congruent to $A'B'C'$, so $BA'' = B'A' = BA$.The angles $BCA$ and $BCA''$ are right angles and together make $180^\circ$, so the points $A$, $C$ and $A''$ lie on one line.Triangle $ABA''$ is isosceles ($BA = BA''$), so $\angle A = \angle A''$.Triangles $ABC$ and $A''BC$ have equal hypotenuses and equal acute angles at $A$ and $A''$, so they’re congruent by hypotenuse and acute angle. And $A''BC$ is congruent to $A'B'C'$, so $ABC$ is congruent to $A'B'C'$.
Experiment 4. Take a stick and break it in two random places. Will the three pieces make a triangle? Before pressing the button, try to guess how often it’ll work.
Each trial is a point of the square: $u$ and $v$ are the places of the two breaks. Green points are successful trials. Press “×1000” a few times.
The share of successful trials creeps towards $25\%$, and that’s no accident.
If a stick is broken at two points chosen randomly, independently and uniformly along its length, the three pieces make a triangle with probability $\frac14$.
Each trial is a point of a square, and the probability is the share of its area where a triangle can be made.
Let the stick have length $1$ and the breaks be at the points $u$ and $v$. The pair $(u, v)$ is a point of the unit square. With a uniform choice, landing in any part of the square is as likely as its area: the probability equals the area. There’s more on computing probabilities like this in chapter 47.First the case $u < v$: the half of the square above the diagonal. The pieces are $u$, $v - u$ and $1 - v$.A triangle can be made when each piece is less than the sum of the other two. All the pieces add up to $1$, so “a piece is less than the sum of the other two” means “a piece is less than $\frac12$”. The conditions $u < \frac12$, $1 - v < \frac12$ and $v - u < \frac12$ cut out a triangle with vertices $(0, \frac12)$, $(\frac12, \frac12)$ and $(\frac12, 1)$. It’s a right triangle with legs $\frac12$, and its area is $\frac12 \cdot \frac12 \cdot \frac12 = \frac18$.The case $u > v$ is symmetric about the diagonal: another triangle just like it, of area $\frac18$.Together, $\frac18 + \frac18 = \frac14$ of the square’s area: that’s the probability.
Area: half the base times the height
An altitude (height) of a triangle is the perpendicular dropped from a vertex to the line containing the opposite side. In an obtuse triangle two altitudes fall on extensions of the sides, outside the triangle.
The area of a triangle is half the product of a side and the altitude to that side: $S = \frac12 a h_a$.
Any side of the triangle; it’s called the base.The altitude to that side. Base and altitude go as a pair: another side has another altitude, but the product stays the same.Example: in the triangle with sides $13$, $14$, $15$ the altitude to side $14$ is $12$, so $S = \frac12 \cdot 14 \cdot 12 = 84$. The altitude to side $13$ is then $\frac{2 \cdot 84}{13} = \frac{168}{13} \approx 12.92$.
Two identical triangles make a parallelogram, and a parallelogram can be re-cut into a rectangle.
Drop from vertex $C$ the altitude $CH$ to side $AB$; its length is $h$. Assume the foot $H$ lies on the segment $AB$.Rotate a copy of the triangle half a turn about the midpoint of side $BC$. The vertices $B$ and $C$ swap, and vertex $A$ goes to a point $D$. A half-turn takes every line to a line parallel to it, so $CD \parallel AB$ and $BD \parallel AC$: the copy together with the original triangle makes the parallelogram $ABDC$. Its area is twice the triangle’s.Cut the right triangle $AHC$ off the parallelogram and slide it along the side by the length $AB$. It fits snugly against side $BD$, which is parallel and equal to side $AC$.We get a rectangle with sides $a$ and $h$; its area is $a \cdot h$ (why the area of a rectangle is the product of its sides is explained in chapter 20). Re-cutting doesn’t change area, so the parallelogram has area $ah$, and the triangle half that: $S = \frac12 a h$.
If the foot of the altitude falls on an extension of the side (the angle at $A$ or $B$ is obtuse), the triangle is completed to a right triangle and the areas of two right triangles are subtracted; the answer is the same.
And if the altitude isn’t known, only the three sides? The area is still determined: by the third test, three sides fix the whole triangle. The formula was found by Heron of Alexandria, a first-century engineer, and written down in his treatise Metrica. According to al-Biruni, Archimedes already knew it.
The area of a triangle with sides $a$, $b$, $c$ is $S = \sqrt{p(p - a)(p - b)(p - c)}$, where $p = \frac{a + b + c}{2}$ is the semiperimeter.
The semiperimeter: $p = \frac{a + b + c}{2}$.How much the semiperimeter exceeds each side. By the triangle inequality all three differences are positive; if one of them is zero, the triangle is flattened and the area is zero.Example: for the sides $13$, $14$, $15$ we get $p = 21$ and $S = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84$, the same as with the altitude. For the Egyptian triangle $3$, $4$, $5$: $p = 6$, $S = \sqrt{6 \cdot 3 \cdot 2 \cdot 1} = 6$.
Find the altitude in terms of the sides, which needs Pythagoras’ theorem from the next chapter (it’s proved without Heron’s formula, so the argument isn’t circular), and then algebra does the rest.
Drop the altitude $h$ to side $c$. Its foot $H$ divides $c$ into the pieces $AH = x$ and $HB = c - x$.By Pythagoras’ theorem in the right triangle $AHC$: $h^2 = b^2 - x^2$.In the right triangle $BHC$: $h^2 = a^2 - (c - x)^2 = a^2 - c^2 + 2cx - x^2$.Equate the two expressions for $h^2$: $b^2 - x^2 = a^2 - c^2 + 2cx - x^2$. The $x^2$ terms cancel, and $x = \frac{b^2 + c^2 - a^2}{2c}$.The area is $S = \frac12 ch$, so $16S^2 = 4c^2h^2 = 4c^2(b^2 - x^2) = (2bc)^2 - (2cx)^2 = (2bc)^2 - (b^2 + c^2 - a^2)^2$.The difference of squares factors: $\bigl(2bc - b^2 - c^2 + a^2\bigr)\bigl(2bc + b^2 + c^2 - a^2\bigr) = \bigl(a^2 - (b - c)^2\bigr)\bigl((b + c)^2 - a^2\bigr)$. Each bracket is again a difference of squares: we get $(a - b + c)(a + b - c)(b + c - a)(b + c + a)$. These brackets are $2(p - b)$, $2(p - c)$, $2(p - a)$ and $2p$, so $16S^2 = 16\,p(p - a)(p - b)(p - c)$. Divide by $16$ and take the square root.
If the foot of the altitude lies on an extension of the side, the number $x$ comes out negative or greater than $c$, but all the equations stay true, and the formula is the same.
Find the area of the triangle with sides $5$, $5$ and $6$.
The semiperimeter is $p = \frac{5 + 5 + 6}{2} = 8$. By Heron’s formula $S = \sqrt{8 \cdot 3 \cdot 3 \cdot 2} = \sqrt{144} = 12$. Check: the altitude to the base $6$ bisects it and equals $4$ (the triangle $3$, $4$, $5$), and $\frac12 \cdot 6 \cdot 4 = 12$.
If other data are known, two sides and an angle, or a side and two angles, the sides and area are found with sines and cosines (chapter 21). The “Solving a triangle” solver does it step by step for any set of data.
Similarity and the pyramid’s shadow
Three angles don’t determine a triangle, but they do determine its shape. You can see it in experiment 1: triangles with angles $50^\circ$, $70^\circ$ and $60^\circ$ differ only in scale.
Similar triangles are triangles whose corresponding angles are equal and whose corresponding sides are proportional: $\frac{a'}{a} = \frac{b'}{b} = \frac{c'}{c} = k$. The number $k$ is called the scale factor.
All of similarity rests on one lemma about parallel lines, and the areas from the previous section help prove it.
If a line parallel to side $BC$ of triangle $ABC$ crosses the sides $AB$ and $AC$ at the points $D$ and $E$, then $\frac{AD}{AB} = \frac{AE}{AC}$.
Compare the areas of triangles with shared altitudes: that’s how Euclid proves this lemma in Book VI of the Elements.
Draw the segments $BE$ and $CD$.Triangles $DBE$ and $DCE$ share the base $DE$, and their altitudes to it are equal: the vertices $B$ and $C$ lie on a line parallel to $DE$ and so are the same distance from it. So these triangles have equal areas.Add triangle $ADE$ to each: $S_{ABE} = S_{ADE} + S_{DBE} = S_{ADE} + S_{DCE} = S_{ACD}$.Triangles $ADE$ and $ABE$ share the vertex $E$, and their bases $AD$ and $AB$ lie on one line. They share the altitude from $E$, so their areas are in the ratio of the bases: $\frac{S_{ADE}}{S_{ABE}} = \frac{AD}{AB}$.In the same way triangles $ADE$ and $ACD$ share the altitude from $D$ to the line $AC$: $\frac{S_{ADE}}{S_{ACD}} = \frac{AE}{AC}$.The denominators $S_{ABE}$ and $S_{ACD}$ are equal (step 3), so the left sides are equal too: $\frac{AD}{AB} = \frac{AE}{AC}$.
The special case where $D$ is the midpoint of $AB$ is known as the midpoint theorem: a line parallel to one side through the midpoint of another side bisects the third side. In many countries the intercept theorem as a whole bears the name of Thales.
Triangles are similar if: 1) two angles of one are equal to two angles of the other; or 2) two sides of one are proportional to two sides of the other and the angles between these sides are equal; or 3) the three sides of one are proportional to the three sides of the other.
Cut from the big triangle a corner congruent to the small one, and prove that this corner is similar to the big triangle.
The first test. Let $\angle A = \angle A'$, $\angle B = \angle B'$ and $A'B' < AB$. If $A'B' = AB$, the triangles are simply congruent by the second congruence test.Lay off on $AB$ the segment $AD = A'B'$ and draw $DE \parallel BC$. Then $\angle ADE = \angle B$: these are corresponding angles for the parallels $DE$, $BC$ and the transversal $AB$.Triangle $ADE$ is congruent to $A'B'C'$ by the second congruence test: $AD = A'B'$, angle $A$ equals $\angle A'$, and $\angle ADE = \angle B = \angle B'$.By the intercept theorem $\frac{AD}{AB} = \frac{AE}{AC}$. Call this number $k$.Draw $EF \parallel AB$ up to its intersection with $BC$. The quadrilateral $DBFE$ is a parallelogram, and opposite sides of a parallelogram are equal (the diagonal $DF$ divides it into two triangles congruent by the second test). So $DE = BF$. By the same lemma for angle $C$, $\frac{CE}{CA} = \frac{CF}{CB}$, hence $\frac{AE}{AC} = \frac{BF}{BC} = \frac{DE}{BC}$. All three ratios of the sides of triangles $ADE$ and $ABC$ equal $k$, and their angles are equal, so $ADE$, and with it $A'B'C'$, is similar to $ABC$.The second test. Let $\frac{A'B'}{AB} = \frac{A'C'}{AC} = k$ and $\angle A = \angle A'$. The same construction gives a triangle $ADE$ with $AD = k \cdot AB = A'B'$, $AE = k \cdot AC = A'C'$ and the same angle $A$. By the first congruence test $ADE$ is congruent to $A'B'C'$, and $ADE$ is similar to $ABC$.The third test. If all three ratios $\frac{A'B'}{AB}$, $\frac{A'C'}{AC}$, $\frac{B'C'}{BC}$ equal $k$, the sides of triangle $ADE$, which are $k \cdot AB$, $k \cdot AC$, $k \cdot BC$, coincide with the sides of $A'B'C'$. By the third congruence test $ADE$ is congruent to $A'B'C'$, and again $A'B'C'$ is similar to $ABC$.
Experiment 5. The Sun is far away, and its rays fall on the ground parallel to one another. So the pyramid with its shadow and the pole with its shadow form similar right triangles: both have a right angle and the same angle of the rays, and by the first test two equal angles are enough. Change the height of the Sun and see what happens.
The dimensions are those of the Great Pyramid of Khufu when it was built: height $146.6$ m, side of the base $230.3$ m. The pole, $2$ m tall, is shown enlarged in the inset.The height of the pyramid, which we want.The height of the pole, which can be measured.The pyramid’s shadow, measured from the point directly under the apex, that is, from the centre of the base. You can’t walk to the centre, so half the side of the base is added to the visible shadow.The pole’s shadow at the same moment.Example: with the Sun at $36^\circ$ the pole’s shadow is $s \approx 2.753$ m, the pyramid’s visible shadow $86.63$ m, $S \approx 86.63 + 115.15 = 201.78$ m, and $H \approx 2 \cdot \frac{201.78}{2.753} \approx 146.6$ m. The formula holds as long as the pyramid’s shadow reaches beyond its base.
The method has an unexpected limitation. The faces of the pyramid slope at about $51.8^\circ$ to the ground. If the Sun is higher than that, the whole shadow falls on the pyramid itself and there’s nothing to measure. At Giza in midsummer the noon Sun rises to $83^\circ$, so Thales would have had to wait for the evening or for winter.
Similarity gives one more fact that we’ll need in the very next section.
A midline (midsegment) of a triangle is the segment joining the midpoints of two of its sides.
A midline of a triangle is parallel to the third side and half its length.
The midline cuts off a corner similar to the whole triangle and half its size.
Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$: $AM = MB$ and $AN = NC$.Triangles $AMN$ and $ABC$ share angle $A$, and the sides on it are proportional: $\frac{AM}{AB} = \frac{AN}{AC} = \frac12$. By the second similarity test these triangles are similar with scale factor $\frac12$.So the third sides are in the same ratio: $MN = \frac12 BC$.Similar triangles have equal corresponding angles: $\angle AMN = \angle ABC$. These are corresponding angles for the lines $MN$, $BC$ and the transversal $AB$, and if corresponding angles are equal, the lines are parallel (chapter 16). So $MN \parallel BC$.
Medians and the centre of mass
A median of a triangle is the segment joining a vertex to the midpoint of the opposite side.
Experiment 6. Draw all three medians. Drag the vertices: do the medians meet at one point, and where on each of them?
The ratios $AG : GM_a$ and the areas of the six parts are recomputed on the fly.
The medians always meet at one point, and that point cuts off a third of each median, counting from the side. Besides, the six triangles the medians cut the figure into all have the same area.
The centroid is the point where the medians of a triangle meet. It’s written $G$ and also called the centre of mass: a uniform triangular plate balances on a point placed there. That it’s the centre of mass was proved by Archimedes in On the Equilibrium of Planes.
The three medians of a triangle meet at one point, which divides each of them in the ratio $2 : 1$, counting from the vertex.
The part of the median from the vertex to the centroid.The part from the centroid to the midpoint of the side: half as long.The whole median from vertex $A$.Example: if a median is $9$, then $AG = 6$ and $GM_a = 3$.
A midline and similarity show where two medians cut each other, and that’s enough to catch the third.
Draw the medians $BM_b$ and $CM_c$; let them meet at the point $G$.The segment $M_cM_b$ joins the midpoints of sides $AB$ and $AC$: it’s a midline, so $M_cM_b \parallel BC$ and $M_cM_b = \frac12 BC$.Triangles $GM_bM_c$ and $GBC$ are similar by two angles: the angles at $G$ are vertical, and $\angle GM_bM_c = \angle GBC$ as alternate angles for the parallels $M_bM_c$, $BC$ and the transversal $BM_b$. The scale factor is $\frac{M_cM_b}{BC} = \frac12$, so $GM_b = \frac12 GB$ and $GM_c = \frac12 GC$.So the median $CM_c$ crosses the median $BM_b$ at the point that divides $BM_b$ in the ratio $2 : 1$ from $B$. The same argument for the medians $AM_a$ and $BM_b$ (with $C$ replaced by $A$) shows that the median $AM_a$ crosses $BM_b$ at the point dividing it $2 : 1$ from $B$.There’s only one such point on the segment $BM_b$. So all three medians pass through $G$, and each is divided by it in the ratio $2 : 1$, counting from the vertex.
The three medians divide a triangle into six triangles of equal area.
Count the areas in pairs, remembering that a median divides a triangle into two halves of equal area.
Triangles $GBM_a$ and $GCM_a$ have equal bases $BM_a = M_aC$ and a shared altitude from $G$, so their areas are equal; call each of them $x$. In the same way we get pairs of equal areas $y$ (at the point $M_b$) and $z$ (at the point $M_c$).The median $AM_a$ divides triangle $ABC$ into two parts of equal area: they have equal bases and a shared altitude from $A$. One part has the areas $z$, $z$ and $x$, the other $x$, $y$ and $y$. From $2z + x = x + 2y$ we get $z = y$.The median $BM_b$, for the same reason, gives $2x + y = y + 2z$, so $x = z$.So $x = y = z$: all six parts have equal area, each one sixth of the triangle.
Bisectors and the incircle
An angle bisector of a triangle is the segment of the bisector of one of its angles from the vertex to the opposite side.
Experiment 7. Draw the three bisectors. They too meet at one point, and that point is the same distance from all three sides.
Dashed: the perpendiculars from the bisectors’ meeting point to the sides. They’re always equal.
The bisectors of a triangle meet at one point. This point is the same distance from all three sides and is the centre of the circle inscribed in the triangle.
The bisector of an angle is the locus of points equidistant from its arms. That’s enough.
Draw the bisectors of angles $A$ and $B$; let them meet at the point $I$. Drop from $I$ the perpendiculars $IP$, $IQ$ and $IR$ to the sides $AB$, $AC$ and $BC$.The point $I$ lies on the bisector of angle $A$. The right triangles $AIP$ and $AIQ$ are congruent by hypotenuse and acute angle: they share the hypotenuse $AI$, and their angles at $A$ are equal. So $IP = IQ$.In the same way, since $I$ lies on the bisector of angle $B$, we get $IP = IR$.So $IQ = IR$. The right triangles $CIQ$ and $CIR$ are congruent by the hypotenuse $CI$ and a leg, so $\angle ICQ = \angle ICR$: the ray $CI$ is the bisector of angle $C$. All three bisectors pass through $I$.The circle with centre $I$ and radius $r = IP = IQ = IR$ passes through three points on the sides. Every other point of a side is farther from $I$ than the foot of the perpendicular: in a right triangle the hypotenuse is longer than a leg, since the larger side is opposite the right angle. So the circle touches each side at exactly one point: it’s inscribed in the triangle.
The incircle is the circle that touches all three sides of the triangle. Its centre $I$, the incentre, is where the bisectors meet, and its radius $r$ is the distance from $I$ to the sides. Every triangle has an incircle, and only one: the centre of any inscribed circle is equidistant from the sides, so it lies on all three bisectors.
$r = \frac{S}{p}$, where $S$ is the area of the triangle and $p$ its semiperimeter.
The radius of the incircle.The area of the triangle.The semiperimeter $p = \frac{a + b + c}{2}$.Example: the triangle $3$, $4$, $5$ has area $6$ and $p = 6$, so $r = 1$.
Cut the triangle into three parts with a common vertex at the incentre.
Join the centre $I$ to the vertices. The triangle falls into three: $AIB$, $BIC$ and $CIA$.The radii to the points of contact are perpendicular to the sides (that’s how we built the incircle), so in each of the three triangles the altitude from vertex $I$ equals $r$.The areas of the three parts are $\frac12 cr$, $\frac12 ar$ and $\frac12 br$.Add them: $S = \frac12 (a + b + c)\, r = pr$, so $r = \frac{S}{p}$.
Find the radius of the circle inscribed in the triangle with sides $13$, $14$ and $15$.
We already know the area: $S = 84$, and the semiperimeter is $p = 21$. So $r = \frac{84}{21} = 4$.
Perpendicular bisectors and the circumcircle
Experiment 8. Draw the perpendicular bisectors of all three sides. Move vertex $C$ through the right-angle position and watch the point where they meet.
The “acute”, “right” and “obtuse” presets put the vertices in typical positions.
We already used the main property of the perpendicular bisector in the last chapter, when we constructed the midpoint of a segment. It’s time to prove it.
A point is equidistant from the ends of a segment if and only if it lies on the perpendicular bisector of the segment.
Both halves of the statement come down to the congruence of two triangles with a shared side.
Let $M$ be the midpoint of segment $AB$, and $m$ the line through $M$ perpendicular to $AB$.If a point $X$ lies on $m$, triangles $XMA$ and $XMB$ are congruent by the first test: the side $XM$ is shared, $MA = MB$, and the angles at $M$ are right angles. So $XA = XB$.Conversely, let $XA = XB$ (with $X$ not equal to $M$). Then triangles $XMA$ and $XMB$ have all three sides equal: $XA = XB$, $MA = MB$, and $XM$ is shared. By the third test they’re congruent, and $\angle XMA = \angle XMB$.These angles are adjacent and together make $180^\circ$, so each is a right angle: $XM \perp AB$, and the point $X$ lies on $m$.
The perpendicular bisectors of the sides of a triangle meet at one point. This point is the same distance from all the vertices and is the centre of the circle circumscribed about the triangle.
The lemma, applied twice, does all the work.
Draw the perpendicular bisectors of sides $AB$ and $BC$. They meet: if they were parallel, the lines $AB$ and $BC$ perpendicular to them would be parallel too, yet they share the point $B$. Call the meeting point $O$.$O$ lies on the perpendicular bisector of $AB$, so $OA = OB$.$O$ lies on the perpendicular bisector of $BC$, so $OB = OC$.So $OA = OC$, and by the lemma the point $O$ lies on the perpendicular bisector of $AC$. All three perpendicular bisectors pass through $O$.The circle with centre $O$ and radius $OA = OB = OC$ passes through all three vertices.
The circumcircle is the circle through all three vertices of the triangle. Its centre $O$, the circumcentre, is where the perpendicular bisectors of the sides meet, and its radius is written $R$.
Where the centre ends up depends on the largest angle: experiment 8 shows that in an acute triangle $O$ is inside, and in an obtuse one outside, beyond the longest side. This observation is easiest to prove with inscribed angles, and we’ll come back to it in chapter 19. But the in-between case, the right triangle, we’ll deal with right now.
The centre of the circle circumscribed about a right triangle is the midpoint of the hypotenuse, and its radius is half the hypotenuse.
Show that the midpoint of the hypotenuse lies on the perpendicular bisector of a leg.
Let angle $C$ be a right angle, and $M$ the midpoint of the hypotenuse $AB$.Draw through $M$ the line parallel to the leg $BC$; it crosses the leg $AC$ at a point $N$. By the intercept theorem $\frac{AN}{AC} = \frac{AM}{AB} = \frac12$, so $N$ is the midpoint of $AC$.$MN \parallel BC$, and $BC \perp AC$, so $MN \perp AC$ too. So $MN$ is the perpendicular bisector of $AC$, and by the lemma $MA = MC$.We’ve got $MA = MB = MC$: the circle with centre $M$ and radius $\frac{AB}{2}$ passes through all three vertices.
Altitudes and the orthocentre
Experiment 9. Now the altitudes. In an obtuse triangle their segments don’t even cross, but the lines they lie on do.
Dashed: the extensions of the sides and altitudes. Turn the triangle from obtuse to acute and back.
The orthocentre is the point where the lines containing the altitudes of a triangle meet. It’s written $H$.
The lines of the altitudes always meet at one point, but proving it is harder than for the medians and bisectors: the Elements has no such theorem. A clever move helps: reduce the altitudes to perpendicular bisectors, about which we already know everything.
The lines containing the altitudes of a triangle meet at one point.
The altitudes of a triangle are the perpendicular bisectors of another triangle, twice as large.
Draw through each vertex of triangle $ABC$ the line parallel to the opposite side. The three lines form a big triangle $A'B'C'$; the vertex $A'$ lies opposite $A$, and so on.The quadrilateral $ABA'C$ is a parallelogram: its opposite sides are parallel by construction. Opposite sides of a parallelogram are equal (the diagonal divides it into two triangles congruent by the second test), so $CA' = AB$. From the parallelogram $ABCB'$ we likewise get $B'C = AB$. So $C$ is the midpoint of side $A'B'$, and in the same way $A$ and $B$ are the midpoints of the other two sides.The altitude from vertex $C$ is perpendicular to $AB$, and so to the side $A'B'$ parallel to it, and passes through its midpoint $C$. It’s the perpendicular bisector of side $A'B'$.In the same way the altitudes from $A$ and $B$ are the perpendicular bisectors of the other two sides of triangle $A'B'C'$.The perpendicular bisectors of the sides of any triangle meet at one point. So the three altitudes of triangle $ABC$ pass through one point $H$.
Where is the orthocentre of a right triangle?
The legs themselves serve as two altitudes: each is perpendicular to the other. They meet at the right-angle vertex, and the third altitude passes through it too. The midpoint of the hypotenuse is home to a different centre: the circumcentre.
The Euler line and the nine-point circle
We have four special points: $G$, $I$, $O$ and $H$. Experiment 10 is the last and the most unexpected. Put the centroid, the circumcentre and the orthocentre on one drawing and move the vertices.
$O$ is the circumcentre, $G$ the centroid, $H$ the orthocentre, $N$ the centre of the nine-point circle. The hollow dot is where the nine-point circle touches the incircle.
The points $O$, $G$ and $H$ always lie on one line, and $G$ divides the segment $OH$ in the ratio $1 : 2$. Leonhard Euler noticed this in 1765.
The Euler line is the line containing the circumcentre $O$, the centroid $G$ and the orthocentre $H$ of any triangle except an equilateral one: in an equilateral triangle all three points coincide.
In any triangle the circumcentre $O$, the centroid $G$ and the orthocentre $H$ lie on one line, with $G$ between $O$ and $H$ and $GH = 2 \cdot OG$.
The distance from the centroid to the orthocentre.The distance from the circumcentre to the centroid. The point $G$ lies between $O$ and $H$.Example: for the triangle with vertices $(0, 0)$, $(6, 0)$ and $(2, 4)$ we get $O = (3, 1)$, $G = (\frac83, \frac43)$, $H = (2, 2)$. Then $OG = \frac{\sqrt2}{3}$ and $GH = \frac{2\sqrt2}{3}$: exactly twice as much.
Shrink the plane towards the point $G$ by half, with a half-turn, and see where the altitudes go.
Consider this transformation of the plane: each point $X$ goes to the point $X'$ on the extension of the segment $XG$ beyond $G$, with $GX' = \frac12 GX$.The centroid divides each median in the ratio $2 : 1$ from the vertex, so the vertices $A$, $B$, $C$ go to the midpoints of the opposite sides $M_a$, $M_b$, $M_c$.Each line goes to a line parallel to it. For any points $X$ and $Y$, triangles $GXY$ and $GX'Y'$ are similar by the second test: the angles at $G$ are vertical, and the sides $GX'$ and $GY'$ are half as long as $GX$ and $GY$. So $\angle GX'Y' = \angle GXY$, and the lines $X'Y'$ and $XY$ are parallel: their alternate angles are equal.The altitude from $A$ passes through $A$ perpendicular to $BC$. Its image passes through $M_a$ and is parallel to it, so it’s also perpendicular to $BC$: it’s the perpendicular bisector of $BC$. So the three altitudes go to the three perpendicular bisectors.The common point of the altitudes, $H$, goes to the common point of the perpendicular bisectors, $O$. By the definition of the transformation, $O$ lies on the extension of the segment $HG$ beyond $G$, and $GO = \frac12 GH$.
That’s not all. The midpoints of the three sides, the feet of the three altitudes, and the midpoints of the three segments joining the orthocentre to the vertices, nine points in all, always lie on one circle, and its centre is on the Euler line too.
The nine-point circle is the circle through the midpoints of the sides of a triangle, the feet of its altitudes, and the midpoints of the segments joining the orthocentre to the vertices.
The midpoints of the sides of a triangle, the feet of its altitudes, and the midpoints of the segments joining the orthocentre to the vertices lie on one circle. Its centre $N$ is the midpoint of the segment $OH$, and its radius is $\frac R2$.
Find two rectangles with a shared diagonal: they’ll have a shared circumcircle. The proof is written for an acute triangle; for an obtuse one it’s the same, only the drawing changes.
Call the midpoints of the sides $M_a$, $M_b$, $M_c$, the midpoints of the segments $AH$, $BH$, $CH$ $K_a$, $K_b$, $K_c$, and the feet of the altitudes $H_a$, $H_b$, $H_c$.The segments $M_cM_b$ and $K_bK_c$ are midlines of triangles $ABC$ and $HBC$; both are parallel to $BC$. The segments $M_cK_b$ and $M_bK_c$ are midlines of triangles $ABH$ and $ACH$; both are parallel to $AH$. The line $AH$ is perpendicular to $BC$, so the quadrilateral $M_cM_bK_cK_b$ is a rectangle.The diagonals of a rectangle are equal and bisect each other (in a parallelogram the diagonals bisect each other, which follows from congruent triangles by the second test; in a rectangle they’re also equal, by the first test). So all four vertices are the same distance from the point $N$ where the diagonals cross: they lie on a circle with centre $N$, and $M_cK_c$ and $M_bK_b$ are its diameters.Likewise the quadrilateral $M_cM_aK_cK_a$ is a rectangle: its sides are parallel to $AC$ and $BH$, and $BH \perp AC$. Its diagonal $M_cK_c$ is the same diameter, so its vertices lie on the same circle too. Now the circle holds six points, and $M_aK_a$ is also a diameter of it.The feet of the altitudes. The angle $K_aH_aM_a$ is a right angle: the point $K_a$ lies on the altitude $AH_a$, and $M_a$ on the side $BC$. The midpoint of the hypotenuse of a right triangle is equidistant from all its vertices (the corollary above), and the midpoint of the hypotenuse $K_aM_a$ is $N$. So $NH_a = NM_a$, and $H_a$ lies on the circle. The same goes for $H_b$ and $H_c$.The centre and radius. Under the transformation from the proof of the Euler line, the segment $AH$ goes to $M_aO$, so $OM_a \parallel AH$ and $OM_a = \frac12 AH = K_aH$, with the segments $K_aH$ and $OM_a$ pointing the same way. So $K_aHM_aO$ is a parallelogram, and its diagonals $K_aM_a$ and $OH$ bisect each other at one point: $N$ is the midpoint of $OH$. And $AK_aM_aO$ is a parallelogram too ($AK_a = \frac12 AH = OM_a$), so $K_aM_a = AO = R$, and the radius of the circle is $\frac R2$.
The nine-point circle touches the incircle of the triangle and its three excircles, the circles that touch one side and the extensions of the other two.
In experiment 10 the point of contact with the incircle is marked with a hollow dot: however you move the vertices, the circles stay tangent.
The idea of the proof, and why the whole of it doesn’t fit here
Two circles touch internally if the distance between their centres equals the difference of their radii, so we need to prove $NI = \frac R2 - r$. The distance between the circumcentre and the incentre is given by Euler’s formula $OI^2 = R^2 - 2Rr$, and the distance $HI$ can also be expressed through $R$, $r$ and the semiperimeter. The point $N$ is the midpoint of side $OH$ of triangle $OIH$, and the length of its median $IN$ can be found from the three sides. Substitution leads exactly to $NI = \frac R2 - r$, but each of these formulas needs a proof of its own, and together the calculations take several pages. The shortest proof uses inversion, a transformation that takes circles to circles; it’s covered in chapter 23.
The laboratory closes with a test series. The trainer has three levels: angles and sides, area, and special points and similarity. Each problem is new, and the solution is shown step by step.
Where next
Among all triangles the right-angled ones hold a special place. In the laboratory you could see it at every turn: in a right triangle the orthocentre sits at a vertex and the circumcentre at the midpoint of the hypotenuse. But its main property links not points but sides. Stretch a knotted rope into a triangle with sides $3$, $4$ and $5$, and the angle between the short sides turns out to be a right angle. And if the legs are $1$ and $1$, how long will the hypotenuse be? The answer is given by the most famous theorem in geometry: chapter 18.